Understanding Combinatorics: Probability of Selecting Balls from a Box

[Nikitin]

Hi. OK in a box there are 6 balls, 2 red ones and 4 blue ones. We take 2 balls out of the box without putting any of them back

1) If I wish to know the probability of selecting 2 blue ones, I just do this: (4above2)/(6above2)=6/15 or 4C2/6C2 or (4*3/2)/(6*5/2)2) BUT, if I wish to know the probability 1 blue and 1 red, I must use the first formula here http://en.wikipedia.org/wiki/Hypergeometric_distribution

(4above1)*(2above1)/(6above2)=8/15=p(1 blue and 1 red ball selected)

WHY can't we use the same logic as in the formula in 1) and do this: (4*2/2)/(6*5/2)=4/15 ?? why is just (4*2)/(6*5/2)=8/15 correct?

I mean, why are the number of relevant outcomes "unorganized" or "combinated" in 2) but in 1) they are "organized" and "permuted" ?

Excuse me for my english. I am hoping somebody can please explain this too me?

 

[Nikitin]

so is my english the reason nobody wants to post a reply? If it helps I am talking about "hypergeometric" probability in 2)

 

[mathman]

Instead of mechanically using combinatorial formulas, you are better of with direct calculation.

For the first problem: Prob(first ball chosen is blue)=2/3, prob(second ball chosen is also blue)=3/5. Therefore prob(both balls are blue)=2/5.

For the second problem, there are two mutually exclusive ways of doing it - blue first and red second or red first and blue second. The first has prob 2/3 x 2/5 = 4/15. The second has prob 1/3 x 4/5 = 4/15. Total prob = 8/15.

 

[Nikitin]

aah thank you, i get it now ! :)

 

[blue_raver22]

Hello,

Thank you for your question. I am happy to provide an explanation for the difference in the two formulas you mentioned.

In the first formula, you are calculating the probability of selecting 2 blue balls out of a total of 6 balls in the box. This is called a combination, where the order of the selected balls does not matter. This is why you are using the combination formula (nCr) to calculate the probability in this case.

However, in the second formula, you are calculating the probability of selecting 1 blue ball and 1 red ball out of a total of 6 balls in the box. In this case, the order of the selected balls does matter, as you are specifically looking for one blue ball and one red ball. This is called a permutation, where the order of the selected items does matter. This is why you are using the permutation formula (nPn) to calculate the probability in this case.

To summarize, the first formula is used when the order of the selected items does not matter, while the second formula is used when the order does matter. I hope this helps to clarify the difference between the two formulas.

Best,